Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

d(x) → e(u(x))
d(u(x)) → c(x)
c(u(x)) → b(x)
v(e(x)) → x
b(u(x)) → a(e(x))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

d(x) → e(u(x))
d(u(x)) → c(x)
c(u(x)) → b(x)
v(e(x)) → x
b(u(x)) → a(e(x))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

C(u(x)) → B(x)
D(u(x)) → C(x)

The TRS R consists of the following rules:

d(x) → e(u(x))
d(u(x)) → c(x)
c(u(x)) → b(x)
v(e(x)) → x
b(u(x)) → a(e(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

Q DP problem:
The TRS P consists of the following rules:

C(u(x)) → B(x)
D(u(x)) → C(x)

The TRS R consists of the following rules:

d(x) → e(u(x))
d(u(x)) → c(x)
c(u(x)) → b(x)
v(e(x)) → x
b(u(x)) → a(e(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

D(u(x)) → C(x)
C(u(x)) → B(x)

The TRS R consists of the following rules:

d(x) → e(u(x))
d(u(x)) → c(x)
c(u(x)) → b(x)
v(e(x)) → x
b(u(x)) → a(e(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 2 less nodes.